Question

The vaklue of $\int \frac{\cos 5x+\cos 4x}{1-2\cos 3x}dx$, is equal to

• A. sinx + sin2x + c
• B. $\sin x-\frac{\sin 2x}{2}+c$
• C. $-\sin x-\frac{\sin 2x}{2}+c$
• D. $\sin x+\frac{1}{2}\sin 2x+c$

Answer 1

The correct option is C $-\sin x-\frac{\sin 2x}{2}+c$

$I=\int \frac{\cos 5x+\cos 4x}{1-2\cos 3x}dx$

$I=\int \frac{2\cos \left(\frac{5x+4x}{2}\right).\cos \left(\frac{5x-4x}{2}\right)}{1-2(2{\cos }^2\frac{3x}{2}-1)}dx$

$I=-\int \frac{2\cos \frac{9x}{2}.\cos \frac{x}{2}}{4{\cos }^2\frac{3x}{2}-3}dx$

$I=-\int \frac{2\cos \frac{9x}{2}.\cos \frac{x}{2}.\cos \frac{3x}{2}}{4{\cos }^3\frac{3x}{2}-3\cos \frac{3x}{2}}dx$ [ Multiplying numerator and denominator by $\cos \frac{3x}{2}$ ]

$I=-\int \frac{2\cos \frac{9x}{2}.\cos \frac{x}{2}.\cos \frac{3x}{2}}{\cos 3.\frac{3x}{2}}dx$ [ Since, $\cos 3x=4{\cos }^{3}x-3\cos x$ ]

$I=-\int 2\cos \frac{3x}{2}.\cos \frac{x}{2}dx$

$I=-\int [\cos (\frac{3x}{2}+\frac{x}{2})+\cos (\frac{3x}{2}-\frac{x}{2})]dx$

$I=-\int (\cos 2x+\cos x)dx$

$I=-\frac{1}{2}\sin 2x-\sin x+c$